1 Introduction

Consider fast homoclinic solutions of the following problem:

u ¨ (t)+q(t) u ˙ (t)a(t)|u(t) | p 2 u(t)+W ( t , u ( t ) ) =0,tR,
(1.1)

where p2, tR, u R N , aC(R,R), W C 1 (R× R N ,R) are not periodic in t, and q:RR is a continuous function and Q(t)= 0 s q(s)ds with

lim | t | + Q(t)=+.
(1.2)

When q(t)0, problem (1.1) reduces to the following special second-order Hamiltonian system:

u ¨ (t)a(t)|u(t) | p 2 u(t)+W ( t , u ( t ) ) =0,tR.
(1.3)

When p=0, problem (1.1) reduces to the following second-order damped vibration problem:

u ¨ (t)+q(t) u ˙ (t)a(t)u(t)+W ( t , u ( t ) ) =0,tR.
(1.4)

If we take p=2 and q(t)0, then problem (1.1) reduces to the following second-order Hamiltonian system:

u ¨ (t)a(t)u(t)+W ( t , u ( t ) ) =0,tR.
(1.5)

The existence of homoclinic orbits plays an important role in the study of the behavior of dynamical systems. If a system has transversely intersected homoclinic orbits, then it must be chaotic. If it has smoothly connected homoclinic orbits, then it cannot stand the perturbation, and its perturbed system probably produces chaotic phenomena. The first work about homoclinic orbits was done by Poincaré [1].

Recently, the existence and multiplicity of homoclinic solutions and periodic solutions for Hamiltonian systems have been extensively studied by critical point theory. For example, see [230] and references therein. In [6, 16, 17], the authors considered homoclinic solutions for the special Hamiltonian system (1.3) in weighted Sobolev space. Later, Shi et al. [31] obtained some results for system (1.3) with a p-Laplacian, which improved and generalized the results in [6, 16, 17]. However, there is little research as regards the existence of homoclinic solutions for damped vibration problems (1.4) when q(t)0. In 2008, Wu and Zhou [32] obtained some results for damped vibration problems (1.4) with some boundary value conditions by variational methods. Zhang and Yuan [33, 34] studied the existence of homoclinic solutions for (1.4) when q(t)c is a constant. Later, Chen et al. [35] investigated fast homoclinic solutions for (1.4) and obtained some new results under more relaxed assumptions on W(t,x), which resolved some open problems in [33]. Zhang [36] obtained infinitely many solutions for a class of general second-order damped vibration systems by using the variational methods. Zhang [37] investigated subharmonic solutions for a class of second-order impulsive systems with damped term by using the mountain pass theorem.

Motivated by [21, 23, 3234, 3842], we will establish some new results for (1.1) in weighted Sobolev space. In order to introduce the concept of fast homoclinic solutions for problem (1.1), we first state some properties of the weighted Sobolev space E on which the certain variational functional associated with (1.1) is defined and the fast homoclinic solutions are the critical points of the certain functional.

Let

X= { u H 1 , 2 ( R , R N ) | R e Q ( t ) [ | u ˙ ( t ) | 2 + | u ( t ) | 2 ] d t < + } ,

where Q(t) is defined in (1.2) and for u,vX, let

u,v= R e Q ( t ) [ ( u ˙ ( t ) , v ˙ ( t ) ) + ( u ( t ) , v ( t ) ) ] dt.

Then X is a Hilbert space with the norm given by

u= ( R e Q ( t ) [ | u ˙ ( t ) | 2 + | u ( t ) | 2 ] d t ) 1 / 2 .

It is obvious that

X L 2 ( e Q ( t ) )

with the embedding being continuous. Here L p ( e Q ( t ) ) (2p<+) denotes the Banach spaces of functions on ℝ with values in R N under the norm

u p = { R e Q ( t ) | u ( t ) | p d t } 1 / p .

If σ is a positive, continuous function on ℝ and 1<s<+, let

L σ s ( e Q ( t ) ) = { u L loc 1 ( e Q ( t ) ) | R σ ( t ) e Q ( t ) | u ( t ) | s d t < + } .

L σ s equipped with the norm

u s , σ = ( R σ ( t ) e Q ( t ) | u ( t ) | s d t ) 1 / s

is a reflexive Banach space.

Set E=X L a p ( e Q ( t ) ), where a is the function given in condition (A). Then E with its standard norm is a reflexive Banach space. Similar to [33, 35], we have the following definition of fast homoclinic solutions.

Definition 1.1 If (1.2) holds, a solution of (1.1) is called a fast homoclinic solution if uE.

The functional φ corresponding to (1.1) on E is given by

φ(u)= R e Q ( t ) [ 1 2 | u ˙ ( t ) | 2 + a ( t ) p | u ( t ) | p W ( t , u ( t ) ) ] dt,uE.
(1.6)

Clearly, it follows from (W1) or (W1)′ that φ:ER. By Theorem 2.1 of [43], we can deduce that the map

ua(t) e Q ( t ) |u(t) | p 2 u(t)

is continuous from L a p ( e Q ( t ) ) in the dual space L a 1 / ( p 1 ) p 1 ( e Q ( t ) ), where p 1 = p p 1 . As the embeddings EX L γ ( e Q ( t ) ) for all γ2 are continuous, if (A) and (W1) or (W1)′ hold, then φ C 1 (E,R) and one can easily check that

φ ( u ) , v = R e Q ( t ) [ ( u ˙ ( t ) , v ˙ ( t ) ) + a ( t ) | u ( t ) | p 2 ( u ( t ) , v ( t ) ) ( W ( t , u ( t ) ) , v ( t ) ) ] d t , u E .
(1.7)

Furthermore, the critical points of φ in E are classical solutions of (1.1) with u(±)=0.

Now, we state our main results.

Theorem 1.1 Suppose that a, q, and W satisfy (1.2) and the following conditions:

  1. (A)

    Let p>2, a(t) is a continuous, positive function onsuch that for all tR

    a(t)α | t | β ,α>0,β>(p2)/2.

(W1) W(t,x)= W 1 (t,x) W 2 (t,x), W 1 , W 2 C 1 (R× R N ,R), and there exists a constant R>0 such that

1 a ( t ) |W(t,x)|=o ( | x | p 1 ) as x0

uniformly in t(,R][R,+).

(W2) There is a constant μ>p such that

0<μ W 1 (t,x) ( W 1 ( t , x ) , x ) ,(t,x)R× R N {0}.

(W3) W 2 (t,0)=0 and there exists a constant ϱ(p,μ) such that

W 2 (t,x)0, ( W 2 ( t , x ) , x ) ϱ W 2 (t,x),(t,x)R× R N .

Then problem (1.1) has at least one nontrivial fast homoclinic solution.

Theorem 1.2 Suppose that a, q, and W satisfy (1.2), (A), (W2), and the following conditions:

(W1)′ W(t,x)= W 1 (t,x) W 2 (t,x), W 1 , W 2 C 1 (R× R N ,R), and

1 a ( t ) |W(t,x)|=o ( | x | p 1 ) as x0

uniformly in tR.

(W3)′ W 2 (t,0)=0 and there exists a constant ϱ(p,μ) such that

( W 2 ( t , x ) , x ) ϱ W 2 (t,x),(t,x)R× R N .

Then problem (1.1) has at least one nontrivial fast homoclinic solution.

Theorem 1.3 Suppose that a, q, and W satisfy (1.2), (A), (W1)-(W3), and the following assumption:

(W4) W(t,x)=W(t,x), (t,x)R× R N .

Then problem (1.1) has an unbounded sequence of fast homoclinic solutions.

Theorem 1.4 Suppose that a, q, and W satisfy (1.2), (A), (W1)′, (W2), (W3)′, and (W4). Then problem (1.1) has an unbounded sequence of fast homoclinic solutions.

Remark 1.1 It is easy to see that our results hold true even if p=2. To the best of our knowledge, similar results for problem (1.1) cannot be seen in the literature, from this point, our results are new. As pointed out in [17], condition (A) can be replaced by more general assumption: a(t)+ as |t|+.

The rest of this paper is organized as follows: in Section 2, some preliminaries are presented. In Section 3, we give the proofs of our results. In Section 4, some examples are given to illustrate our results.

2 Preliminaries

Let E and be given in Section 1, by a similar argument in [41], we have the following important lemma.

Lemma 2.1 For any uE,

u 1 2 e 0 u= 1 2 e 0 { R e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 ,
(2.1)
| u ( t ) | { t + e Q ( s ) e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 | u ( t ) | 1 e 0 4 { t + e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2
(2.2)

and

| u ( t ) | { t e Q ( s ) e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 1 e 0 4 { t e Q ( s ) [ | u ˙ ( s ) | 2 + | u ( s ) | 2 ] d s } 1 / 2 ,
(2.3)

where u =ess sup t R |u(t)|, e 0 = e min { Q ( t ) : t R } .

The following lemma is an improvement result of [16].

Lemma 2.2 If a satisfies assumption (A), then

the embedding  L a p ( e Q ( t ) ) L 2 ( e Q ( t ) )  is continuous.
(2.4)

Moreover, there exists a Hilbert space Z such that

the embeddings  L a p ( e Q ( t ) ) Z L 2 ( e Q ( t ) )  are continuous;
(2.5)
the embedding XZ L 2 ( e Q ( t ) )  is compact.
(2.6)

Proof Let θ=p/(p2), θ =p/2, we have

u 2 2 = R e Q ( t ) | u ( t ) | 2 d t = R a 1 / θ a 1 / θ e Q ( t ) / θ e Q ( t ) / θ | u ( t ) | 2 d t ( R a θ / θ e Q ( t ) d t ) 1 / θ ( R a e Q ( t ) | u ( t ) | 2 θ d t ) 1 / θ = a 1 ( R a e Q ( t ) | u ( t ) | p d t ) 2 / p = a 1 u p , a 2 ,

where from (A) and (1.2), a 1 = ( R a 2 / ( p 2 ) e Q ( t ) d t ) ( p 2 ) / p <+. Then (2.4) holds.

By (A), there exists a continuous positive function ρ such that ρ(t)+ as |t|+ and

a 2 = ( R ρ θ a θ / θ e Q ( t ) d t ) 1 / θ <+.

Since

u 2 , ρ 2 = R ρ e Q ( t ) | u ( t ) | 2 d t = R ρ a 1 / θ a 1 / θ e Q ( t ) / θ e Q ( t ) / θ | u ( t ) | 2 d t ( R ρ θ a θ / θ e Q ( t ) d t ) 1 / θ ( R a e Q ( t ) | u ( t ) | p d t ) 1 / θ = a 2 u p , a 2 ,

(2.5) holds by taking Z= L ρ 2 ( e Q ( t ) ).

Finally, as XZ is the weighted Sobolev space Γ 1 , 2 (R,ρ,1), it follows from [43] that (2.6) holds. □

The following two lemmas are the mountain pass theorem and the symmetric mountain pass theorem, which are useful in the proofs of our theorems.

Lemma 2.3 [44]

Let E be a real Banach space and I C 1 (E,R) satisfying (PS)-condition. Suppose I(0)=0 and:

  1. (i)

    There exist constants ρ,α>0 such that I B ρ ( 0 ) α.

  2. (ii)

    There exists an eE B ¯ ρ (0) such that I(e)0.

Then I possesses a critical value cα which can be characterized as

c= inf h Φ max s [ 0 , 1 ] I ( h ( s ) ) ,

where Φ={hC([0,1],E)|h(0)=0,h(1)=e} and B ρ (0) is an open ball in E of radius ρ centered at 0.

Lemma 2.4 [44]

Let E be a real Banach space and I C 1 (E,R) with I even. Assume that I(0)=0 and I satisfies (PS)-condition, assumption (i) of Lemma  2.3 and the following condition:

  1. (iii)

    For each finite dimensional subspace E E, there is r=r( E )>0 such that I(u)0 for u E B r (0), B r (0) is an open ball in E of radius r centered at 0.

Then I possesses an unbounded sequence of critical values.

Lemma 2.5 Assume that (W2) and (W3) or (W3)′ hold. Then for every (t,x)R× R N ,

  1. (i)

    s μ W 1 (t,sx) is nondecreasing on (0,+);

  2. (ii)

    s ϱ W 2 (t,sx) is nonincreasing on (0,+).

The proof of Lemma 2.5 is routine and we omit it.

3 Proofs of theorems

Proof of Theorem 1.1 Step 1. The functional φ satisfies the (PS)-condition. Let { u n }E satisfying φ( u n ) is bounded and φ ( u n )0 as n. Then there exists a constant C 1 >0 such that

|φ( u n )| C 1 , φ ( u n ) E μ C 1 .
(3.1)

From (1.6), (1.7), (3.1), (W2), and (W3), we have

2 C 1 + 2 C 1 u n 2 φ ( u n ) 2 μ φ ( u n ) , u n = μ 2 μ u ˙ n 2 2 + 2 R e Q ( t ) [ W 2 ( t , u n ( t ) ) 1 μ ( W 2 ( t , u n ( t ) ) , u n ( t ) ) ] d t 2 R e Q ( t ) [ W 1 ( t , u n ( t ) ) 1 μ ( W 1 ( t , u n ( t ) ) , u n ( t ) ) ] d t + ( 2 p 2 μ ) R a ( t ) e Q ( t ) | u n ( t ) | p d t μ 2 μ u ˙ n 2 2 + ( 2 p 2 μ ) u n p , a p .

It follows from Lemma 2.2, μ>p>2, and the above inequalities that there exists a constant C 2 >0 such that

u n C 2 ,nN.
(3.2)

Now we prove that u n u 0 in E. Passing to a subsequence if necessary, it can be assumed that u n u 0 in E. Since Q(t) as |t|, we can choose T>R such that

Q(t)ln ( C 2 2 ξ 2 ) for |t|T.
(3.3)

It follows from (2.2), (3.2), and (3.3) that

| u n ( t ) | 2 t + e Q ( s ) e Q ( s ) [ | u ˙ n ( s ) | 2 + | u n ( s ) | 2 ] d s ξ 2 C 2 2 u n 2 ξ 2 for  t T  and  n N .
(3.4)

Similarly, by (2.3), (3.2), and (3.3), we have

| u n (t) | 2 ξ 2 for tT and nN.
(3.5)

Since u n u 0 in E, it is easy to verify that u n (t) converges to u 0 (t) pointwise for all tR. Hence, it follows from (3.4) and (3.5) that

| u 0 (t)|ξfor t(,T][T,+).
(3.6)

Since e Q ( t ) e 0 >0 on [T,T]=J, the operator defined by S:EX(J):uu | J is a linear continuous map. So u n u 0 in X(J). The Sobolev theorem implies that u n u 0 uniformly on J, so there is n 0 N such that

T T e Q ( t ) |W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) || u n (t) u 0 (t)|dt<εfor n n 0 .
(3.7)

For any given number ε>0, by (W1), we can choose ξ>0 such that

|W(t,x)|εa(t) | x | p 1 for |t|R and |x|ξ.
(3.8)

From (3.8), we have

e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | 2 e Q ( t ) [ ε a ( t ) ( | u n ( t ) | p 1 + | u 0 ( t ) | p 1 ) ] 2 e Q ( t ) [ ε 2 p 1 a ( t ) | u n ( t ) u 0 ( t ) | p 1 + ε ( 1 + 2 p 1 ) a ( t ) | u 0 ( t ) | p 1 ] 2 2 2 p ε 2 a 2 ( t ) e Q ( t ) | u n ( t ) u 0 ( t ) | 2 ( p 1 ) + ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 a 2 ( t ) e Q ( t ) | u 0 ( t ) | 2 ( p 1 ) : = g n ( t ) .
(3.9)

Moreover, since a(t) is a positive continuous function on ℝ and u n (t) converges to u 0 (t) pointwise for all tR, it follows from (3.9) that

lim n g n (t)= ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 a 2 (t) e Q ( t ) | u 0 (t) | 2 ( p 1 ) :=g(t)for a.e. tR

and

lim n R ( T , T ) g n ( t ) d t = lim n R ( T , T ) e Q ( t ) [ 2 2 p ε 2 a 2 ( t ) | u n ( t ) u 0 ( t ) | 2 ( p 1 ) + ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 a 2 ( t ) | u 0 ( t ) | 2 ( p 1 ) ] d t = 2 2 p ( ε ) 2 lim n R ( T , T ) a 2 ( t ) e Q ( t ) | u n ( t ) u 0 ( t ) | 2 ( p 1 ) d t + ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 R ( T , T ) a 2 ( t ) e Q ( t ) | u 0 ( t ) | 2 ( p 1 ) d t = ( 2 ε ) 2 ( 1 + 2 p 1 ) 2 R ( T , T ) a 2 ( t ) e Q ( t ) | u 0 ( t ) | 2 ( p 1 ) d t = R g ( t ) d t < + .

From Lebesgue’s dominated convergence theorem, (3.4), (3.5), (3.6), (3.9), and the above inequalities, we have

lim n R ( T , T ) e Q ( t ) |W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | 2 dt=0.
(3.10)

From Lemma 2.2, we have u n u 0 in L 2 ( e Q ( t ) ). Hence, by (3.10),

R ( T , T ) e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | | u n ( t ) u 0 ( t ) | d t ( R ( T , T ) e Q ( t ) | W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) | 2 d t ) 1 / 2 × ( R ( T , T ) e Q ( t ) | u n ( t ) u 0 ( t ) | 2 d t ) 1 / 2

tends to 0 as n+, which together with (3.7) shows that

R e Q ( t ) |W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) || u n (t) u 0 (t)|dt0as n.
(3.11)

From (1.7), we have

0 φ ( u n ) φ ( u 0 ) , u n u 0 = u ˙ n u ˙ 0 2 2 + R a ( t ) e Q ( t ) ( | u n ( t ) | p 2 u n ( t ) | u 0 ( t ) | p 2 u 0 ( t ) ) ( u n ( t ) u 0 ( t ) ) d t R e Q ( t ) ( W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) , u n ( t ) u 0 ( t ) ) d t u ˙ n u ˙ 0 2 2 + C 3 R a ( t ) e Q ( t ) ( | u n ( t ) u 0 ( t ) | p ) d t R e Q ( t ) ( W ( t , u n ( t ) ) W ( t , u 0 ( t ) ) , u n ( t ) u 0 ( t ) ) d t , n ,
(3.12)

where C 3 is a positive constant. It follows from (3.11) and (3.12) that

u ˙ n 2 u ˙ 0 2 as n
(3.13)

and

R a(t) e Q ( t ) | u n (t) | p dt R a(t) e Q ( t ) | u 0 (t) | p dtas n.
(3.14)

Hence, u n u 0 in E by (3.13) and (3.14). This shows that φ satisfies (PS)-condition.

Step 2. From (W1), there exists δ(0,1) such that

|W(t,x)| 1 2 a(t) | x | p 1 for |t|R,|x|δ.
(3.15)

By (3.15) and W(t,0)=0, we have

|W(t,x)| 1 2 p a(t) | x | p for |t|R,|x|δ.
(3.16)

Let

C 4 =sup { W 1 ( t , x ) a ( t ) | t [ R , R ] , x R , | x | = 1 } .
(3.17)

Set σ=min{1/ ( 2 p C 4 + 1 ) 1 / ( μ p ) ,δ} and u= 2 e 0 σ:=ρ, it follows from Lemma 2.1 that |u(t)|σδ<1 for tR. From Lemma 2.5(i) and (3.17), we have

R R e Q ( t ) W 1 ( t , u ( t ) ) d t { t [ R , R ] : u ( t ) 0 } e Q ( t ) W 1 ( t , u ( t ) | u ( t ) | ) | u ( t ) | μ d t C 4 R R a ( t ) e Q ( t ) | u ( t ) | μ d t C 4 σ μ p R R a ( t ) e Q ( t ) | u ( t ) | p d t 1 2 p R R a ( t ) e Q ( t ) | u ( t ) | p d t .
(3.18)

By (W3), (3.16), and (3.18), we have

φ ( u ) = 1 2 R e Q ( t ) | u ˙ ( t ) | 2 d t + R a ( t ) p e Q ( t ) | u ( t ) | p d t R e Q ( t ) W ( t , u ( t ) ) d t = 1 2 u ˙ 2 2 + 1 p u p , a p R ( R , R ) e Q ( t ) W ( t , u ( t ) ) d t R R e Q ( t ) W ( t , u ( t ) ) d t 1 2 u ˙ 2 2 + 1 p u p , a p R ( R , R ) e Q ( t ) W ( t , u ( t ) ) d t R R e Q ( t ) W 1 ( t , u ( t ) ) d t 1 2 u ˙ 2 2 + 1 p u p , a p 1 2 p R ( R , R ) a ( t ) e Q ( t ) | u ( t ) | p d t 1 2 p R R a ( t ) e Q ( t ) | u ( t ) | p d t = 1 2 u ˙ 2 2 + 1 2 p u p , a p .

Therefore, we can choose a constant α>0 depending on ρ such that φ(u)α for any uE with u=ρ.

Step 3. From Lemma 2.5(ii) and (2.1), we have for any uE

3 3 e Q ( t ) W 2 ( t , u ( t ) ) d t = { t [ 3 , 3 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t + { t [ 3 , 3 ] : | u ( t ) | 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t { t [ 3 , 3 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) | u ( t ) | ) | u ( t ) | ϱ d t + 3 3 e Q ( t ) max | x | 1 W 2 ( t , x ) d t u ϱ 3 3 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + 3 3 e Q ( t ) max | x | 1 W 2 ( t , x ) d t ( 1 2 e 0 ) ϱ u ϱ 3 3 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + 3 3 e Q ( t ) max | x | 1 W 2 ( t , x ) d t = C 5 u ϱ + C 6 ,
(3.19)

where C 5 = ( 1 2 e 0 ) ϱ 3 3 e Q ( t ) max | x | = 1 W 2 (t,x)dt, C 6 = 3 3 e Q ( t ) max | x | 1 W 2 (t,x)dt. Take ωE such that

|ω(t)|= { 1 for  | t | 1 , 0 for  | t | 3 ,
(3.20)

and |ω(t)|1 for |t|(1,3]. For s>1, from Lemma 2.5(i) and (3.20), we get

1 1 e Q ( t ) W 1 ( t , s ω ( t ) ) dt s μ 1 1 e Q ( t ) W 1 ( t , ω ( t ) ) dt= C 7 s μ ,
(3.21)

where C 7 = 1 1 e Q ( t ) W 1 (t,ω(t))dt>0. From (W3), (1.6), (3.19), (3.20), and (3.21), we get for s>1

φ ( s ω ) = s 2 2 ω ˙ 2 2 + s p p ω p , a p + R e Q ( t ) [ W 2 ( t , s ω ( t ) ) W 1 ( t , s ω ( t ) ) ] d t s 2 2 ω ˙ 2 2 + s p p ω p , a p + 3 3 e Q ( t ) W 2 ( t , s ω ( t ) ) d t 1 1 e Q ( t ) W 1 ( t , s ω ( t ) ) d t s 2 2 ω ˙ 2 2 + s p p ω p , a p + C 5 s ϱ ω ϱ + C 6 C 7 s μ .
(3.22)

Since μ>ϱ>p and C 7 >0, it follows from (3.22) that there exists s 1 >1 such that s 1 ω>ρ and φ( s 1 ω)<0. Set e= s 1 ω(t), then eE, e= s 1 ω>ρ, and φ(e)=φ( s 1 ω)<0. It is easy to see that φ(0)=0. By Lemma 2.3, φ has a critical value c>α given by

c= inf g Φ max s [ 0 , 1 ] φ ( g ( s ) ) ,
(3.23)

where

Φ= { g C ( [ 0 , 1 ] , E ) : g ( 0 ) = 0 , g ( 1 ) = e } .

Hence, there exists u E such that

φ ( u ) = c , φ ( u ) = 0 .

The function u is the desired solution of problem (1.1). Since c>0, u is a nontrivial fast homoclinic solution. The proof is complete. □

Proof of Theorem 1.2 In the proof of Theorem 1.1, the condition W 2 (t,x)0 in (W3) is only used in the proofs of (3.2) and Step 2. Therefore, we only need to prove that (3.2) and Step 2 still hold if we use (W1)′ and (W3)′ instead of (W1) and (W3). We first prove that (3.2) holds. From (W2), (W3)′, (1.6), (1.7), and (3.1), we have

2 C 1 + 2 C 1 μ ϱ u n 2 φ ( u n ) 2 ϱ φ ( u n ) , u n = ϱ 2 ϱ u ˙ n 2 2 + 2 R e Q ( t ) [ W 2 ( t , u n ( t ) ) 1 ϱ ( W 2 ( t , u n ( t ) ) , u n ( t ) ) ] d t 2 R e Q ( t ) [ W 1 ( t , u n ( t ) ) 1 ϱ ( W 1 ( t , u n ( t ) ) , u n ( t ) ) ] d t + 2 ( 1 p 1 ϱ ) R a ( t ) e Q ( t ) | u n ( t ) | p d t ϱ 2 ϱ u ˙ n 2 2 + 2 ( 1 p 1 ϱ ) u n p , a p ,

which implies that there exists a constant C 3 >0 such that (3.2) holds. Next, we prove that Step 2 still holds. From (W1)′, there exists δ(0,1) such that

|W(t,x)| 1 2 a(t) | x | p 1 for tR,|x|δ.
(3.24)

By (3.24) and W(t,0)=0, we have

|W(t,x)| 1 2 p a(t) | x | p for tR,|x|δ.
(3.25)

Let u= 2 e 0 δ:=ρ, it follows from Lemma 2.1 that |u(t)|δ. It follows from (1.6) and (3.25) that

φ ( u ) = 1 2 R e Q ( t ) | u ˙ ( t ) | 2 d t + R a ( t ) e Q ( t ) p | u ( t ) | p d t R e Q ( t ) W ( t , u ( t ) ) d t 1 2 u ˙ 2 2 + 1 p u p , a p R 1 2 p a ( t ) e Q ( t ) | u ( t ) | p d t = 1 2 u ˙ 2 2 + 1 2 p u p , a p .

Therefore, we can choose a constant α>0 depending on ρ such that φ(u)α for any uE with u=ρ. The proof of Theorem 1.2 is complete. □

Proof of Theorem 1.3 Condition (W4) shows that φ is even. In view of the proof of Theorem 1.1, we know that φ C 1 (E,R) and satisfies (PS)-condition and assumption (i) of Lemma 2.3. Now, we prove that (iii) of Lemma 2.4. Let E be a finite dimensional subspace of E. Since all norms of a finite dimensional space are equivalent, there exists d>0 such that

ud u .
(3.26)

Assume that dim E =m and { u 1 , u 2 ,, u m } is a basis of E such that

u i =d,i=1,2,,m.
(3.27)

For any u E , there exists λ i R, i=1,2,,m such that

u(t)= i = 1 m λ i u i (t)for tR.
(3.28)

Let

u = i = 1 m | λ i | u i .
(3.29)

It is easy to see that is a norm of E . Hence, there exists a constant d >0 such that d u u. Since u i E, by Lemma 2.1, we can choose R 1 >R such that

| u i (t)|< d δ 1 + m ,|t|> R 1 ,i=1,2,,m,
(3.30)

where δ is given in (3.25). Let

Θ= { i = 1 m λ i u i ( t ) : λ i R , i = 1 , 2 , , m ; i = 1 m | λ i | = 1 } = { u E : u = d } .
(3.31)

Hence, for uΘ, let t 0 = t 0 (u)R such that

|u( t 0 )|= u .
(3.32)

Then by (3.26)-(3.29), (3.31), and (3.32), we have

d = u u d d d u = d d | u ( t 0 ) | = d d | i = 1 m λ i u i ( t 0 ) | d i = 1 m | λ i | | u i ( t 0 ) | , u Θ .
(3.33)

This shows that |u( t 0 )| d and there exists i 0 {1,2,,m} such that | u i 0 ( t 0 )| d /m, which together with (3.30), implies that | t 0 | R 1 . Let R 2 = R 1 +1 and

γ=min { e Q ( t ) W 1 ( t , x ) : R 2 t R 2 , d 2 | x | d 2 e 0 } .
(3.34)

Since W 1 (t,x)>0 for all tR and x R N {0}, and W 1 C 1 (R× R N ,R), it follows that γ>0. For any uE, from Lemma 2.1 and Lemma 2.5(i), we have

R 2 R 2 e Q ( t ) W 2 ( t , u ( t ) ) d t = { t [ R 2 , R 2 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t + { t [ R 2 , R 2 ] : | u ( t ) | 1 } e Q ( t ) W 2 ( t , u ( t ) ) d t { t [ R 2 , R 2 ] : | u ( t ) | > 1 } e Q ( t ) W 2 ( t , u ( t ) | u ( t ) | ) | u ( t ) | ϱ d t + R 2 R 2 e Q ( t ) max | x | 1 W 2 ( t , x ) d t u ϱ R 2 R 2 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + R 2 R 2 e Q ( t ) max | x | 1 W 2 ( t , x ) d t ( 1 2 e 0 ) ϱ u ϱ R 2 R 2 e Q ( t ) max | x | = 1 W 2 ( t , x ) d t + R 2 R 2 e Q ( t ) max | x | 1 W 2 ( t , x ) d t = C 8 u ϱ + C 9 ,
(3.35)

where C 8 = ( 1 2 e 0 ) ϱ R 2 R 2 e Q ( t ) max | x | = 1 W 2 (t,x)dt, C 9 = R 2 R 2 e Q ( t ) max | x | 1 W 2 (t,x)dt. Since u ˙ i L 2 ( e Q ( t ) ), i=1,2,,m, it follows that there exists ε(0,( ( d ) 2 e 0 )/(32 m 2 d 2 )) such that

t + ε t ε | u ˙ i ( s ) | d s = t + ε t ε e Q ( s ) 2 e Q ( s ) 2 | u ˙ i ( s ) | d s 1 e 0 t + ε t ε e Q ( s ) 2 | u ˙ i ( s ) | d s 1 e 0 ( 2 ε ) 1 / 2 ( t + ε t ε e Q ( s ) | u ˙ i ( s ) | 2 d s ) 1 / 2 ( 2 ε e 0 ) 1 / 2 u ˙ i 2 d 4 m for  t R , i = 1 , 2 , , m .
(3.36)

Then for uΘ with |u( t 0 )|= u and t[ t 0 ε, t 0 +ε], it follows from (3.28), (3.31), (3.32), (3.33), and (3.36) that

| u ( t ) | 2 = | u ( t 0 ) | 2 + 2 t 0 t ( u ˙ ( s ) , u ( s ) ) d s | u ( t 0 ) | 2 2 t 0 ε t 0 + ε | u ( s ) | | u ˙ ( s ) | d s | u ( t 0 ) | 2 2 | u ( t 0 ) | t 0 ε t 0 + ε | u ˙ ( s ) | d s | u ( t 0 ) | 2 2 | u ( t 0 ) | i = 1 m | λ i | t 0 ε t 0 + ε | u ˙ i ( s ) | d s ( d ) 2 2 .
(3.37)

On the other hand, since ud for uΘ, then

|u(t)| u d 2 e 0 ,tR,uΘ.
(3.38)

Therefore, from (3.34), (3.37), and (3.38), we have

R 2 R 2 e Q ( t ) W 1 ( t , u ( t ) ) dt t 0 ε t 0 + ε e Q ( t ) W 1 ( t , u ( t ) ) dt2εγfor uΘ.
(3.39)

By (3.30) and (3.31), we have

|u(t)| i = 1 m | λ i || u i (t)|δfor |t| R 1 ,uΘ.
(3.40)

By (1.6), (3.16), (3.35), (3.39), (3.40), and Lemma 2.5, we have for uΘ and r>1

φ ( r u ) = r 2 2 u ˙ 2 2 + r p p u p , a p + R e Q ( t ) [ W 2 ( t , r u ( t ) ) W 1 ( t , r u ( t ) ) ] d t r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ R e Q ( t ) W 2 ( t , u ( t ) ) d t r μ R e Q ( t ) W 1 ( t , u ( t ) ) d t = r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ R ( R 2 , R 2 ) e Q ( t ) W 2 ( t , u ( t ) ) d t r μ R ( R 2 , R 2 ) e Q ( t ) W 1 ( t , u ( t ) ) d t + r ϱ R 2 R 2 e Q ( t ) W 2 ( t , u ( t ) ) d t r μ R 2 R 2 e Q ( t ) W 1 ( t , u ( t ) ) d t r 2 2 u ˙ 2 2 + r p p u p , a p r ϱ R ( R 2 , R 2 ) e Q ( t ) W ( t , u ( t ) ) d t r μ R 2 R 2 e Q ( t ) W 1 ( t , u ( t ) ) d t + r ϱ R 2 R 2 e Q ( t ) W 2 ( t , u ( t ) ) d t r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ 2 p R ( R 2 , R 2 ) a ( t ) e Q ( t ) | u ( t ) | p d t + r ϱ ( C 8 u ϱ + C 9 ) 2 ε γ r μ r 2 2 u ˙ 2 2 + r p p u p , a p + r ϱ 2 p u p , a p + r ϱ ( C 8 u ϱ + C 9 ) 2 ε γ r μ r 2 2 d 2 + r p p d p + r ϱ 2 p d p + C 8 ( r d ) ϱ + C 9 r ϱ 2 ε γ r μ .
(3.41)

Since μ>ϱ>p>2, we deduce that there exists r 0 = r 0 (d, d , C 8 , C 9 , R 1 , R 2 ,ε,γ)= r 0 ( E )>1 such that

φ(ru)<0for uΘ and r r 0 .

It follows that

φ(u)<0for u E  and ud r 0 ,

which shows that (iii) of Lemma 2.4 holds. By Lemma 2.4, φ possesses an unbounded sequence { c n } n = 1 of critical values with c n =φ( u n ), where u n is such that φ ( u n )=0 for n=1,2, . If { u n } is bounded, then there exists C 10 >0 such that

u n C 10 for nN.
(3.42)

In a similar fashion to the proof of (3.4) and (3.5), for the given δ in (3.16), there exists R 3 >R such that

| u n (t)|δfor |t| R 3 ,nN.
(3.43)

Hence, by (1.6), (2.1), (3.16), (3.42), and (3.43), we have

1 2 u ˙ n 2 2 + 1 p u n p , a p = c n + R e Q ( t ) W ( t , u n ( t ) ) d t = c n + R [ R 3 , R 3 ] e Q ( t ) W ( t , u n ( t ) ) d t + R 3 R 3 e Q ( t ) W ( t , u n ( t ) ) d t c n 1 2 p R [ R 3 , R 3 ] a ( t ) e Q ( t ) | u n ( t ) | p d t R 3 R 3 e Q ( t ) | W ( t , u n ( t ) ) | d t c n 1 2 p u n p , a p R 3 R 3 e Q ( t ) max | x | 2 e 0 C 10 | W ( t , x ) | d t ,

which, together with (3.42), implies that

c n 1 2 u ˙ n 2 2 + 3 2 p u n p , a p + R 3 R 3 max | x | 2 e 0 C 10 e Q ( t ) |W(t,x)|dt<+.

This contradicts the fact that { c n } n = 1 is unbounded, and so { u n } is unbounded. The proof is complete. □

Proof of Theorem 1.4 In view of the proofs of Theorem 1.2 and Theorem 1.3, the conclusion of Theorem 1.4 holds. The proof is complete. □

4 Examples

Example 4.1 Consider the following system:

u ¨ (t)+t u ˙ (t)a(t)|u(t) | 1 / 2 u(t)+W ( t , u ( t ) ) =0,a.e. tR,
(4.1)

where q(t)=t, p=5/2, tR, u R N , aC(R,(0,)), and a satisfies (A). Let

W(t,x)=a(t) ( i = 1 m a i | x | μ i j = 1 n b j | x | ϱ j ) ,

where μ 1 > μ 2 >> μ m > ϱ 1 > ϱ 2 >> ϱ n >5/2, a i , b j >0, i=1,,m, j=1,,n. Let

W 1 (t,x)=a(t) i = 1 m a i | x | μ i , W 2 (t,x)=a(t) j = 1 n b j | x | ϱ j .

Then it is easy to check that all the conditions of Theorem 1.3 are satisfied with μ= μ m and ϱ= ϱ 1 . Hence, problem (4.1) has an unbounded sequence of fast homoclinic solutions.

Example 4.2 Consider the following system:

u ¨ (t)+ ( t + t 3 ) u ˙ (t)a(t)|u(t) | 4 u(t)+W ( t , u ( t ) ) =0,a.e. tR,
(4.2)

where q(t)=t+ t 3 , p=6, tR, u R N , aC(R,(0,)), and a satisfies (A). Let

W(t,x)=a(t) [ a 1 | x | μ 1 + a 2 | x | μ 2 b 1 ( cos t ) | x | ϱ 1 b 2 | x | ϱ 2 ] ,

where μ 1 > μ 2 > ϱ 1 > ϱ 2 >6, a 1 , a 2 >0, b 1 , b 2 >0. Let

W 1 (t,x)=a(t) ( a 1 | x | μ 1 + a 2 | x | μ 2 ) , W 2 (t,x)=a(t) [ b 1 ( cos t ) | x | ϱ 1 + b 2 | x | ϱ 2 ] .

Then it is easy to check that all the conditions of Theorem 1.4 are satisfied with μ= μ 2 and ϱ= ϱ 1 . Hence, by Theorem 1.4, problem (4.2) has an unbounded sequence of fast homoclinic solutions.